Half Life Problems Worksheet Answers

One-half-Life or previously known every bit the Half-Life Period is ane of the common terminologies used in Science to describe the radioactivity of a item sample or chemical element within a certain period of time.

All the same, this concept is also widely used to describe various types of decay processes, especially exponential and non-exponential decay. Apart from science, the term is used in medical sciences to correspond the biological half-life of certain chemicals in the man torso or in drugs.

Definition: Half-Life is unremarkably divers every bit the time a radioactive substance (or ane half the atoms) needs to disintegrate or transform into a dissimilar substance. The principle was commencement discovered in 1907 past Ernest Rutherford. Information technology is normally represented by the symbol Ug or t_1/two.

Information technology can also be referred to equally the fourth dimension taken for half of the reactions to consummate or the time at which the concentration of the reactant is reduced to one-half of its original value is chosen the half-life period of the reaction.

Half-Life Chemistry Questions with Solutions

Q1. An isotope of caesium (Cs-137) has a half-life of 30 years. If 1.0g of Cs-137 disintegrates over a menstruum of 90 years, how many grams of Cs-137 would remain?

a.) i.25 g

b.) 0.125 g

c.) 0.00125 thousand

d.) 12.5 thou

Correct Answer- (b.) 0.125 thousand

Q2. Selenium-83 has a half-life of 25.0 minutes. How many minutes would information technology take for a x.0 mg sample to decay and simply have 1.25 mg of it remain?

a.) 75 minutes

b.) 75 days

c.) 75 seconds

d.) 75 hours

Right Respond- (a.) 75 minutes

Q3. How long does it take a 100.00g sample of Every bit-81, with a one-half-life of 33 seconds, to decay to vi.25g?

a.) 122 seconds

b.) 101 seconds

c.) 132 seconds

d.) 22 seconds

Correct Answer– (c.) 132 seconds

Q4. What is the one-half-life of a radioactive isotope if a 500.0g sample decays to 62.5g in 24.3 hours?

a.) 8.1 hours

b.) half dozen.1 hours

c.) 5 hours

d.) 24 hours

Correct Answer- (a.) eight.i hours

Q5. What is the half-life of Polonium-214 if, after 820 seconds, a ane.0g sample decays to 0.03125g?

a.) 164 minutes

b.) 164 seconds

c.) 64 seconds

d.) 160 minutes

Correct Respond- (b.) 164 seconds

Q6. The half-life of Zn-71 is 2.4 minutes. If one had 100.0 g at the beginning, how many grams would exist left after 7.2 minutes have elapsed?

Reply.

To brainstorm, we'll count the number of one-half-lives that have passed. This tin exist obtained by doing the following:

One-half-life (t½) = two.iv mins

Time (t) = 7.2 mins

Number of half-lives

Number of one-half-lives

\(\begin{array}{fifty}n=\frac{t}{t_{1/2}}\cease{array} \)

n = 7.2/2.iv = 3

Thus, three half-lives have passed.

Finally, we volition calculate the remaining amount. This tin be obtained past doing the post-obit:

N₀ (original amount) = 100 g

(north) = number of half-lives

Amount remaining (N) =?

\(\begin{assortment}{l}N = \frac{N_{0}}{N^{n}}\cease{assortment} \)

N = 100 / 2³

N = 100 / 8

Northward = 12.five thousand

As a result, the amount of Zn-71 remaining afterward seven.ii minutes is 12.5 thou.

Q7. Pd-100 has a half-life of 3.six days. If one had 6.02 x 10²³ atoms at the start, how many atoms would be present later 20.0 days?

Answer.

Half-life = 3.6 days

Initial atoms = vi.02 ×10²³ atoms

Time = 20days

To summate the atoms nowadays afterwards xx days, we utilise the formula beneath.

\(\begin{array}{fifty}N = N_{0}\times \frac{1}{2}\times \frac{t}{t_{1/2}}\end{assortment} \)

\(\begin{assortment}{l}Due north = half-dozen.02\times 10^{23}\times \frac{1}{2}\times \frac{twenty}{3.half-dozen}=1.28\times x^{22}\end{array} \)

Thus, the number of atoms available is 1.28 × ten²²atoms.

Q8. Os-182 has a half-life of 21.5 hours. How many grams of a x.0 gram sample would have decayed after exactly 3 half-lives?

Answer. The amount of the radioactive substance that will remain subsequently 3- half- lives=(½)³ × a,

where a = initial concentration of the radioactive element.

a= 10 g

So, amount of the radioactive substance that remains aftet iii- half-lives=( ½)³x10 = 10/eight= 1.25 thousand.

Therefore, the number of grams of the radioactive substance that rust-covered in 3 half-lives = (10 – 1.25) g

= 8.75 g

Q9. After 24.0 days, 2.00 milligrams of an original 128.0 milligram sample remain. What is the half-life of the sample?

Answer. The remaining decimal fraction is:

2.00 mg / 128.0 mg = 0.015625

The half-lives that must take expired to go to 0.015625?

(½)ⁿ = 0.015625

n log 0.5 = 0.015625

n = log 0.5 / 0.015625 due north = half-dozen

Calculation of the one-half-life:

24 days divided by 6 half-lives equals 4.00 days

Q10. A radioactive isotope decayed to 17/32 of its original mass after lx minutes. Find the one-half-life of this radioisotope.

Answer. The amount that remains

17/32 = 0.53125

(1/two)^northward = 0.53125

n log 0.five = log 0.53125

n = 0.91254

One-half-lives that have elapsed are therefore, n = 0.9125

60 minutes divided by 0.91254 equals 65.75 minutes.

Therefore, due north = 66 minutes

Q11. How long will information technology take for a 40 gram sample of I-131 (one-half-life = 8.040 days) to decay to i/100 of its original mass?

Answer. (i/2)^northward = 0.01

n log 0.5 = log 0.01

north = 6.64

vi.64 10 eight.040 days = 53.four days

Therefore, it volition take 53.4 days to disuse to 1/100 of its original mass.

Q12. At time nil, at that place are ten.0 grams of West-187. If the half-life is 23.9 hours, how much will exist present at the finish of one day? Two days? Seven days?

Respond.

24.0 hr / 23.ix hr/half-life = one.0042 half-lives

One day = one one-half-life; (ane/2)^1.0042 = 0.4985465 remaining = 4.98 g

2 days = two half-lives; (one/2)^ii.0084 = 0.2485486 remaining = 2.48 thousand

7 days = 7 half-lives; (1/two)^vii.0234 = 0.0076549 remaining = 0.0765 g

Q13. 100.0 grams of an isotope with a half-life of 36.0 hours is present at time zip. How much time volition accept elapsed when 5.00 grams remains?

Answer.

The afraction amount remaining will be-

5.00 / 100.0 = 0.05

(1/2)^north = 0.05

n log 0.five = log 0.05

n = 4.32 one-half-lives

36.0 hours 10 4.32 = 155.half dozen hours

Q14. How much time will be required for a sample of H-iii to lose 75% of its radioactive decay? The half-life of tritium is 12.26 years.

Reply.

If you lot lose 75%, so 25% remains.

(1/ii)ⁿ = 0.25

north = two (Since, (i/two)ⁱⁱ = 1/four and 1/iv = 0.25)

12.26 10 ii = 24.52 years

Therefore, 24.52 years of fourth dimension volition be required for a sample of H-3 to lose 75% of its radioactivity

Q15. The one-half-life for the radioactive disuse of ^xivC is 5730 years. An archaeological artifact containing wood had simply 80% of the ¹⁴C establish in a living tree. Gauge the age of the sample.

Answer. Decay constant, k = 0.693/t_ane/2 = 0.693/5730 years = i/209 × 10^–4/twelvemonth

\(\brainstorm{assortment}{l}t=\frac{two.303}{1000}log\frac{\left [ R \correct ]_{0}}{\left [ R \right ]}\end{array} \)

\(\begin{array}{l}t=\frac{2.303}{1.209\times x^{-iv}}log\frac{100}{fourscore}\cease{array} \)

= 1846 years (approx)

Do Questions on Half-Life

Q1. A newly prepared radioactive nuclide has a decay abiding λ of 10^–vi s^–1. What is the approximate half-life of the nuclide?

a.) 1 hour

b.) i-day

c.) ane calendar week

d.) 1 month

Q2. If the decay constant of a radioactive nuclide is 6.93 x 10^–3 sec^–1, its half-life in minutes is:

a.) 100

b.) 1.67

c.) 6.93

d.) 50

Q3. A beginning-guild reaction takes 40 min for thirty% decomposition. Summate t_1/2.

Q4. What volition exist the fourth dimension for 50% completion of a showtime-order reaction if information technology takes 72 min for 75% completion?

Q5. How much time will it accept for 90% completion of a reaction if 80% of a first-guild reaction was completed in 70 min?

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